codeforces 722E Research Rover
2018-01-16 22:28
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codeforces 722E Research Rover
题意
\(1e5*1e5\)的棋盘中有\(2000\)个坏点,初始给定一个值\(s(1<=s<=1e6)\)。从棋盘左上角走到右下角,只允许向右或者向下走,每经过一个坏点,\(s=\lceil{s/2}\rceil\),求到达棋盘右下角时,\(s\)的值的期望。题解
经过\(log\)级别的坏点之后,\(s\)的值就衡为\(1\)了。对所有坏点根据\(x\)从小到大,\(y\)从小到大排序。\(f_{i, j}\)表示从坏点\(i\)走到右下角经过\(j\)个坏点(不包括\(i\))。
\(f_{i, 0}=Paths(i, 右下角)(所有情况)-\Sigma_{k!=i}{(Paths(i, k)*f_{k, 0})}(经过坏点个数大于零个的情况)\)
\(f_{i, j}=Paths(i, 右下角)(所有情况)-\Sigma_{k!=i}{(Paths(i, k)*f_{k, j})}(经过坏点个数大于j个的情况)-\Sigma_{k=0}^{j-1}{f_{i,k}}(经过坏点个数小于j个的情况)\)
代码
#include<bits/stdc++.h> using namespace std; #define fi first #define se second #define mp make_pair #define pb push_back #define rep(i, a, b) for(int i=(a); i<(b); i++) #define sz(x) (int)x.size() #define de(x) cout<< #x<<" = "<<x<<endl #define dd(x) cout<< #x<<" = "<<x<<" " typedef long long ll; typedef pair<int, int> pii; typedef vector<int> vi; //------ const int N=2020, mod=1e9+7; int n,m,k,s; bool isa; pii a ; ll f [22], jc[201010], inv[201010]; int val[22]; ll kpow(ll a,ll b) { ll res=1; while(b) { if(b&1) res=res*a%mod; a=a*a%mod; b>>=1; } return res; } void init() { jc[0]=1; rep(i,1,201010) jc[i]=jc[i-1]*i%mod; inv[201010-1]=kpow(jc[201010-1], mod-2); for(int i=201010-2;i>=0;--i) inv[i]=inv[i+1]*(i+1)%mod; } ll C(int n, int m) { if(n<0||m<0) return 0; return jc *inv[m]%mod*inv[n-m]%mod; } ll pa(int n, int m) { return C(n+m, m); } void upd(ll &a, ll b) { a=(a+b)%mod; if(a<0) a+=mod; } int main() { init(); while(~scanf("%d%d%d%d",&n,&m,&k,&s)) { ///init isa=0; ///read rep(i,0,k) { int x,y;scanf("%d%d",&x,&y); if(x==1&&y==1) isa=1; a[i]=mp(x, y); } ///get val int p; for(p=0;s>1;++p) { val[p]=s; s=(s+1)/2; } val[p]=1; ///solve if(!isa) a[k++]=mp(1, 1); sort(a,a+k); for(int i=k-1;i>=0;--i) { f[i][0]=pa(n-a[i].fi, m-a[i].se); rep(j,i+1,k) { upd(f[i][0], -pa(a[j].fi-a[i].fi, a[j].se-a[i].se)*f[j][0]%mod); } } rep(j,1,p) { for(int i=k-1;i>=0;--i) { f[i][j]=pa(n-a[i].fi, m-a[i].se); rep(_,i+1,k) { upd(f[i][j], -pa(a[_].fi-a[i].fi, a[_].se-a[i].se)*f[_][j]%mod); } rep(_,0,j) { upd(f[i][j], -f[i][_]); } } } ll cnt=pa(n-1, m-1), ans=kpow(cnt, mod-2), sum=0; rep(j,0,p) { upd(cnt, -f[0][j]); int t=j+isa; upd(sum, f[0][j]*val[t]%mod); } upd(sum, cnt); ans=ans*sum%mod; printf("%lld\n",ans); } return 0; }
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